Question

$ABCD$ is a parallelogram, and $AE$ and $CF$ bisects $\angle A$ and $\angle C$ respectively. Prove that $AE\parallel FC$.

Answer 1

In a parallelogram $ABCD$,

$\angle A=\angle C$ (opposite angles of a parallelogram)

$⟹\frac{1}{2}\angle A=\frac{1}{2}\angle C$ ............(i)

Since $AE$ and $CF$ bisects $\angle A$ and $\angle C$

$⟹\angle DAE=\angle EAB$ and $\angle BCF=\angle FCD$ ................(ii)

From (i) & (ii)

$\angle EAB=\angle FCD$

and $\angle CFB=\angle FCD$ (AB // CD and CF is a transversal)

$\therefore \angle CFB=\angle EAB$ ( corresponding angles )

Hence $AE\parallel FC$